Let $f:(0,1) \rightarrow(0,1)$ be a differentiable function such that $f^{\prime}(x) \neq 0 \forall x \in(0,1)$ and $f\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$. Suppose for all $x$, $\mathop {\lim }\limits_{t \to x} \frac{\int_0^t \sqrt{1-(f(s))^2} d s-\int_0^x \sqrt{1-(f(s))^2} d s}{f(t)-f(x)}=f(x)$. Then the value of $f\left(\frac{1}{4}\right)$ belongs to
$\{\sqrt{7}, \sqrt{6}\}$
$\left\{\frac{\sqrt{7}}{2}, \frac{\sqrt{15}}{2}\right\}$
$\left\{\frac{\sqrt{7}}{4}, \frac{\sqrt{15}}{4}\right\}$
$\left\{\frac{\sqrt{7}}{3}, \frac{\sqrt{15}}{3}\right\}$
If ' $a$ ' is an integer lying in $[-5,30]$, then the probability that the graph of $y=x^2+2(a+4) x-5 a+64$ lies above the $x-$ axis is
$\frac{1}{6}$
$\frac{7}{36}$
$\frac{2}{9}$
$\frac{3}{5}$
Consider a square $A B C D$ of diagonal length 2a. The square is folded along the diagonal $A C$ so that the plane of $\triangle A B C$ is perpendicular to the plane of $\triangle A D C$. In this case the shortest distance between $A B$ and $C D$ is
$\frac{2 a}{\sqrt{3}}$
$\frac{a}{2 \sqrt{3}}$
$\frac{\mathrm{a}}{\sqrt{3}}$
$\frac{\sqrt{3} a}{2}$
If $\int \frac{\left(1-x^2\right)}{\sqrt{x} \sqrt{\left(1+x^2\right)^3}}=\alpha \frac{x^\beta}{\left(1+x^2\right)^\gamma}+C ; \alpha, \beta, \gamma \in \mathbb{R}$ and $C$ is constant of integration, then $\alpha: \beta: \gamma$ will be
$4: 1: 1$
$2: 2: \frac{1}{2}$
$\frac{1}{6}: 2: \frac{1}{2}$
$1: 2: \frac{1}{2}$
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